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Chemistry · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 3 · Q23
Calculate Δ_r G^° and log K_C of the reaction. 3 2 Cr ( s )+3 Cd^2+( aq ) → 2 Cr^3+( aq )+3 Cd ( s ) Given E_Cr^3+ / Cr^°=-0.74 V E_Cd^2+ /…
Calculate $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ and $\displaystyle \log \mathrm{K}_{\mathrm{C}}$ of the reaction. $\displaystyle 3$ \[2 \mathrm{Cr}(\mathrm{~s})+3 \mathrm{Cd}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Cd}(\mathrm{~s}) \] Given $\displaystyle \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V} \] $\displaystyle \left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$
Marking-scheme solution
Eocell= Eocathode−Eoanode
= -$\displaystyle 0.40$-(-$\displaystyle 0.74$)
= +$\displaystyle 0.34$ V
△G0 =-nF Eocell
= -($\displaystyle 6$ x $\displaystyle 96500$ X $\displaystyle 0.34$) J
= - $\displaystyle 196860$ J/ mol
△G0=-$\displaystyle 2.303$ RT logKc
log Kc = ($\displaystyle 196860$)/ $\displaystyle 2.303$ x 8.314x298
=$\displaystyle 34.576$
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CBSE Class 12 Chemistry past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.